User:Egm6321.f10.team1.allison

Problem 1: Derive
This solution is based on the one previously found by Vasquez et al. The primary improvement is the inclusion of a background section, where discussion of why these equations are important is included.

Background
A very common use for the time derivative is in consideration of applied forces. Newton's 2nd law is based in an inertial frame, whereas, whereas most physical systems studied are referenced to a non-inertial coordinate system (the rotating earth). This gives rise to several inertial forces, which go to zero when an inertial reference frame is considered. Amongst these are the Coriolis force and the centrifugal force.

The Coriolis force is calculated as

$${{F}_{c}}=-2m\Omega \times v$$

where m represents the mass of the body, $${\Omega}$$ is the angular velocity of the rotating coordinate system, and v is the linear velocity of the body referenced to the rotating frame. The Coriolis effect of the Earth's rotation can often be ignored, but there are certain engineering applications. Long range ballistic missiles tend to veer off from their intended path if the Coriolis effect is not considered. The Earth's Coriolis effect also causes certain types of waves, the rotation of cyclones, and creates the jet streams and western boundary currents in the oceans. Various other systems may have important Coriolis contributions including rolling jets and rotating missile/gun turrets.

Another important force caused by a rotating frame of reference is the centrifugal force. This force can be defined mathematically as:

$$F=-m\Omega \times (\Omega \times r)$$

where $$ \Omega $$ is defined as above and r is the radius from the center of rotation of the frame. The centrifugal force acts outward from the center of rotation. Some examples of use for this force are: slings, artificial gravity through rotation, spin casting where liquid metal is placed inside of a tube and then the tube spun until the metal is dry thus creating a hollow metal cylinder, and centrifuges to separate materials based on their mass.

Given
The equation

$$f({{y}^{1}}(t),t)$$.

can be used to model the motion of a high speed train.

Find

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$$\frac{\partial }{\partial t}f({{y}^{1}}(t),t)=f{{,}_{s}}{{y}^{1}}{{,}_{t}}+f{{,}_{t}}$$

(p1-3.3)
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$$\frac{d{{t}^{2}}}({{y}^{1}},t)=f{{,}_{s}}{{y}^{1}}{{,}_{tt}}+f{{,}_{ss}}{{\left( {{y}^{1}}{{,}_{t}} \right)}^{2}}+2f{{,}_{st}}{{y}^{1}}{{,}_{t}}+f{{,}_{tt}}$$

(p1-3.4)
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Equation 3
First, for compactness define:


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$$s:={{y}^{1}}(t)$$ (1.1) so that the given term can be re-written as:
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$$f(s,t)$$

(1.2)
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Next, take a derivative with respect to time of this term. This gives the equation:

$$\frac{\partial }{\partial t}f(s,t)=\frac{\partial f}{\partial s}\frac{\partial s}{\partial t}+\frac{\partial f}{\partial t}\frac{\partial t}{\partial t}$$


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$$\frac{\partial }{\partial t}f(s,t)=\frac{\partial f}{\partial s}\frac{\partial s}{\partial t}+\frac{\partial f}{\partial t}$$

(1.3)
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Then substituting back in the definition of s (1):

$$ \frac{d}{dt}f(y^1(t),t)= \frac{\partial f(y^1(t),t)}{\partial s} \frac{\partial y^1}{\partial t}+ \frac{\partial f(y^1(t),t)}{\partial t} $$

Finally, using the notation:

$${{y}^{1}}{{,}_{t}}=\frac{d}{dt}{{y}^{1}}(t)$$

The final equation can be expressed as:


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$$\frac{\partial }{\partial t}f({{y}^{1}}(t),t)=f{{,}_{s}}{{y}^{1}}{{,}_{t}}+f{{,}_{t}}$$
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(p1-3.3)
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Equation 4
Begin with equation 1.3:

$$\frac{\partial }{\partial t}f(s,t)=\frac{\partial f}{\partial s}\frac{\partial s}{\partial t}+\frac{\partial f}{\partial t}$$

Then, taking the time derivative term-wise will produce:

$$\frac{\partial }{\partial t}\left( \frac{\partial f}{\partial t} \right)=\frac{{{\partial }^{2}}f}{\partial {{s}^{2}}}\frac{\partial s}{\partial t}\frac{\partial s}{\partial t}+\frac{{{\partial }^{2}}f}{\partial t\partial s}\frac{\partial s}{\partial t}\frac{\partial t}{\partial t}+\frac{\partial f}{\partial s}\frac{\partial {{s}^{2}}}{\partial t\partial s}+\frac{\partial f}{\partial s}\frac{{{\partial }^{2}}s}{\partial {{t}^{2}}}\frac{\partial t}{\partial t}$$

$$\frac{{{\partial }^{2}}f}{\partial {{t}^{2}}}=\frac{{{\partial }^{2}}f}{\partial {{s}^{2}}}\frac{\partial s}{\partial t}\frac{\partial s}{\partial t}+\frac{{{\partial }^{2}}f}{\partial t\partial s}\frac{\partial s}{\partial t}+\frac{\partial f}{\partial s}+\frac{\partial f}{\partial s}\frac{{{\partial }^{2}}s}{\partial {{t}^{2}}}$$

and again substituting in (1) and switching to comma notation will give the final equation:


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$$\frac{d^2f}{dt^2}=f,_s(y^1,t) \frac{d^2y^1}{dt^2}+f,_{ss}\left(\frac{dy^1}{dt} \right)^2+2f,_{st} \frac{dy^1}{dt}+f,_{tt} $$ (p1-3.4)
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Equation 1

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$${{c}_{o}}({{y}^{1}},t)=-{{F}^{1}}[1-\overline{R}({{u}^{2}}{{,}_{ss}})({{y}^{1}},t)-{{F}^{2}}{{u}^{2}}{{,}_{s}}-\frac{T}{R}+M[[1-\overline{R}({{u}^{2}}{{,}_{ss}})][{{u}^{1}}{{,}_{tt}}-\overline{R}({{u}^{2}}{{,}_{stt}})]+{{u}^{2}}{{,}_{s}}{{u}^{2}}{{,}_{tt}}]$$     (p2-2.1)
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Problem 3: Prove
Prove that the equation:

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$${{c}_{3}}({{y}^{1}},t)\frac{d{{t}^{2}}}$$

(p*-*.*)
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is nonlinear with respect to $$ y^1 $$.

Problem 4: Boundary Value Problem
Solve the equation:

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$$y(x)=cy_{H}^{1}(x)+dy_{H}^{2}+{{y}_{p}}(x)$$

(p*-*.*) using the boundary values:
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$$\begin{align} & y(a)=\alpha \\ & y(b)=\beta \\ \end{align}$$.

Problem 5: Prove
Show that:

$${{L}_{2}}(y_{H}^{1})={{L}_{2}}(y_{H}^{2})=0$$.

$$\begin{align} & {{y}_{1}}=x \\ & {{y}_{1}}'=1 \\ & {{y}_{1}}''=0 \\ \end{align}$$

plugging these into the equation:

$$(1-{{x}^{2}})y''-2xy'+2y=0$$

results in:

$$\begin{align} & (1-{{x}^{2}})[0]-2x[1]+2[x]=0 \\ & -2x+2x=0 \\ & 0=0 \\ \end{align}$$

similarly, the second solution

$$y=\frac{x}{2}\ln \left( \frac{1+x}{1-x}-1 \right)$$

can be shown to be a homogeneous solution to the solution by: